Question

Find the first three terms of the sequence:

Answer 1

replace n by 1, then by 2, then by 3
that is all

Answer 2

to get 2/3 for the first term

Answer 3

@Sparklestaraa

Answer 4

no

Answer 5

its 2/3 or -2/3

Answer 6

the first term is not 2/3 because \(-1)^1=-1

Answer 7

for the first term

Answer 8

then its -2/3

Answer 9

\[(-1)^1=-1\] so the first term is \(-\frac{2}{3}\)

Answer 10

the the second term would be 4/9

Answer 11

@satellite73

Answer 12

also i got -8/27 for the third term? is that right?

Answer 13

@SolomonZelman

Answer 14

@welshfella

Answer 15

oh thank you for coming

Answer 16

its -2/3 ........ 4/9 ....... 8/27 i think

Answer 17

\(\large\color{slate}{ \displaystyle a_n=(-1)\times \left(\frac{2}{3}\right)^n }\)

Answer 18

this is in a form of a(n abstract) exponential function \(\large\color{slate}{ \displaystyle y=a(b)^x }\)

where you start from \(\large\color{slate}{ a }\), multiply (the a) by \(\large\color{slate}{ b }\) each time as \(\large\color{slate}{ \displaystyle x }\) goes up by 1.

(we will disregard all non-natural number values of x, because the number terms in the sequence is natural numbers)

Answer 19

ok

Answer 20

in your form I think you are plugging n=1 for 1st term
n=2 for second term, n=3 for 3rd term and on....

the first term is correct

Answer 21

the 2nd and 3rd are a little off

Answer 22

negative times positive = negative ?

Answer 23

so -4/9 and

Answer 24

and -8/27

Answer 25

OK, your common ratio, as you see is 2/3.
your first term is -2/3, then how 2nd and 3rd can be positive?

Answer 26

all negatives

Answer 27

and?

Answer 28

so -2/3 ...... -4/9 ..... -8/27 one two three terms!

Answer 29

yes

Answer 30

AWESOME ! your the best help I"ve ever recieved!

Answer 31

but, are you sure it isn't
\(\LARGE \color{slate}{ \displaystyle a_n=\left(a_1\right)\times \left(\frac{2}{3}\right)^{n\color{red}{-1}} }\)

rather that it is just
\(\LARGE \color{slate}{ \displaystyle a_n=\left(a_1\right)\times \left(\frac{2}{3}\right)^{n} }\)

??

Answer 32

sure?

Answer 33

well it is but i didnt type that

Answer 34

it is negative though

Answer 35

yes, so we did everything fine if it is \(\LARGE \color{slate}{ \displaystyle a_n=\left(a_1\right)\times \left(\frac{2}{3}\right)^{n} }\)

Answer 36

would it be \(\LARGE \color{slate}{ \displaystyle a_n=\left(a_1\right)\times \left(\frac{2}{3}\right)^{n\color{red}{-1}} }\)

then \(\LARGE \color{slate}{ a_1=-1 }\) (and only then -2/3, -4/9 ... etc)

Answer 37

in any case though you are welcome !

Answer 38

I gotcha..